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Question

In each of the following find the equation of the hyperbola satisfying the given conditions.:

(i)vertices(±2,0),foci(±3,0)

(ii)vertices(0,±,5),foci(0,±,8)

(iii)vertices(0,±,3),foci(0,±,5)

(iv)vertices(±5,0),transverseaxis=8

(v)foci(0,±13),conjugateaxis=24

(vi)foci(±35),thelatusrectum=8

(vii)foci(±,4,0),thelatusrectum=12

(viii)vertices(0,±6),e=53.

(ix)foci(0,±10),passingthrough(2,3)

(x)foci(0,±12),latusrectum=36

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Solution

(i) Since,vertics(±2,0) and foci (±,0) lies on X-axis,as coefficient of Y-axis is zero.

Hence,equation of hyperbola will be of the form

x2a2y2b2=1 ...(i) Where it is given that

a=2 and c=3

c2=a2+b2

9=4+b2

b2=5

Put the values of a2=4 and b2=5 in Eq,(i),we get

x24y25=1

(ii) Since,vertics(0,±5) and foci (0,±8) lies on Y-axis,as coefficient of X-axis is zero.

Hence,equation of hyperbola will be of the form

x2a2y2b2=1 ...(i) Where it is given that (0,±a)=(0,±5)and foci(0,±c)=(0,±8)

a=5 and c=8

c2=a2+b264=25+b2

b2=6425 b2=39

Put a2=25 and b2=39 in Eq,(i),we get

y225x239=1

(iii) Since,vertics(0,±3) and foci (0,±5) lies on Y-axis,as coordinate of X-axis is zero.

Hence,equation of hyperbola will be of the form

x2a2y2b2=1 ...(i) Where it is given that (0,±3)=(0,±a)and foci(0,±5)=(0,±c)

a=3 and c=5

c2=a2+b2

b2=6425

25=9+b2b2=16

Put a2=9 and b2=16 in Eq,(i),we get

y29x216=1

(iv)Since,foci (±5,0)lies on X-axis,as coordinate of Y is zero.

Hence,equation of hyperbola wil be of the form

x2a2y2b2=1 ...(i)

Where it is given that foci(±5,0)=(±c,0)and length of transverse axis is 2a=8

c=5 and 2a=8a=4

c2=a2+b2

25=16+b2b2=9

Put a2=16 and b2=9 in Eq.(i),we get

x216y29=1

(v)Since,foci (0,±13)lies on Y-axis,as coordinate of X is zero.

Hence,equation of hyperbola wil be of the form

y2a2x2b2=1 ...(i)

Where it is given that foci(0,±13)=(0,±c)and conjugaete length 2b=24

c=13 and b=12

c2=a2+b2

169=a2+144a2=169144

a2=25

Put a2=25 and b2=144 in Eq.(i),we get

y225x2144=1

(vi)Since,foci (±35,0)lies on X-axis,as y-coordinate is zero.

Hence,equation of hyperbola wil be of the form

y2a2x2b2=1 ...(i)

Where it is given that foci(±35,0)=(±c,0)and length oflatusrectum

2b2a=8

b2=4a

c2=a2+b2

(35)2=a2+(4a)

9×5=a2+4a

a2+4a45=0

On spliting the middle term

a2+9a5a45=0

a(a+9)5(a+9)=0

(a5)(a+9)=0

a5=0,a+9=0

a=5,9

a cant be negative.

So,a=5b2=4×5=20

Put a2=25 and b2=20 in Eq.(i),we get

x225y220=1

(vii)Since foci(±4,0)lies on X-axis,Hence,equation,of hyperbola will be of the form

y2a2x2b2=1 ...(i)

Where it is given that foci(±4,0)=(±c,0)and length of latusrectum,

\(\frac{2b^2}{a}=12\\)

c=4 and b2=6a ...(ii)

c2=a2+b2

16=a2+6a

a2+6a16=0

Splitting the middle term

a2+8a2a16=0

a(a(a+8)2(a+8)=0

(a2)(a+8)=0

a2=0,a+8=0a=2,8

a cant be negative. So,a=2

b2=6a=6×2=12

Put a2=4 and b2=12 in Eq.(i),we get

x24y212=1

(viii)Let the equation of the hyperbola be

x2a2+y2b2=1

Then vertices=(0,±b)=(0,±6)

b=6 and e=54

e=1+a2b2259=1+a236

2599=a23616=a24+48

So,the equation of hyperbola is,

x248+y236=1y236x248=1

Foci=(0,±be)=(0,±53×6)=(0,±10)

Since,foci(0,±10)lies on Y-axis,as x-coordinate is zero.

Hence,equation of hyperbola will be of the form

y2a2x2b2=1 ...(i)

Where it is given that foci(0,±10)=(0,±c)c=10 Also hyperbola passes through the point (1,3)i.e.,pont~will~satisfy~Eq.(i).

(3)2a2(2)2b2=1

9a24b2=1 ...(ii)

c=10

c2=10

a2+b2=10 (c2=a2+b2)

b2=10a2 ...(iii)

From Ep.(ii),9a2410a2=1

909a24aa2(10a2)=1

9013a2=10a2a4

a423a2+90=0

a418a25a2+90=0

a2(a218)5(a218)=0

a2=5 or a2=18

b2=105=5 If a2=18

b2=1018=8

Which in not possible.

So,a2=b2=5

Put a2=b2=5 in Eq.(i)we get

y25x25=1y2x2=5

(x)Since foci are(0,±12),it follows that c=12 Length of the latus rectum=2b2a=36 or b2=18a

Therefore c2=a2+b2,gives

144=a2+18a i.e.,a2+18a144=0,

So,a=24,6

Sine a cannot be negative,we take a=6 and sob2=108

Therefore,the equation of the required

hyperbola isy236x2108=1,

i.e.,3y2x2=108


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